A member runs from the origin to the point (1.0, 1.0, 0.0) m, at 45 degrees in the horizontal XY plane. Its stiffness is formed in its own local axes and then turned into global axes. Does the solver find the right direction for the member, and is the transformed stiffness matrix still symmetric?

1.Inputs.

Illustrative values, taken from no standard and chosen so the arithmetic can be followed by hand. Most do not describe a real pipe; read them in any consistent set of units.

2.Method.

The local x axis is the vector from start node to end node, divided by its length. Its components are the direction cosines. A correct rotation of a symmetric stiffness matrix leaves it symmetric.

3.Results.

The tests check that:

• The solver’s local x axis has global X and Y components equal to the hand value.
• The transformed global stiffness matrix is symmetric.
• The two recorded components are equal to each other.

What it shows. Compared with an independent hand calculation.

Path exercised. The benchmark calls the solver’s components directly: elements, loads, frame solver and stress recovery. It does not go through the program’s own model-to-solve path.

Agreement. Each computed value must match the reference within an absolute difference of 1.0 × 10⁻⁹ in the case’s own units; counts and structural outcomes must match exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/mechanics/Cargo.toml inclined_member_transform_fixture_checks_direction_cosines_and_symmetry

Hand calculation: validation/hand_calcs/mechanics/inclined_member_transform.md. Test record, with the recorded run of 2026-07-10: mech-inclined-member-transform.md.